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Catapult Experiment

by Sorry!
Tags: catapult, experiment
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Sorry!
#1
Jun10-08, 03:22 PM
P: 571
The culminating task for my physics course was to build a catapult that can launch a ball bearing 2-7m. I built the catapult out of wood and its on a pivot and i'm using a bungee cord to make it snap back when its pulled back.

How to I develop a formula to calculate prior to launch how far it can go based on how far is pull back on the bungee cord? what tests should i conduct.

I am supposed to be able to develop the formula and the teacher gives me a distance i find out how to calibrate my catapult to hit that distance.

I was thinking about just using conservation of energy but i'm not sure about how to go about it since idk the coefficient of elastic nrg for the bungee cord any suggestions?
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tiny-tim
#2
Jun10-08, 03:56 PM
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Hi Sorry!!

You could find the coefficient of stretchiness (I forget what it's called officially ) by measuring how much the bungee stretches when you put known weights on it.
Barny
#3
Jun10-08, 04:01 PM
P: 49
Sounds like fun!

I'm not sure you'll be able to develop a very accurate formula for calculating the range - but using some theory and some tests I reckon you'd be able to get ball park values.

I think using energy conservation would be the easiest bet if you wanted to come up with a formula based around theory, i.e using Hooke's law ect ect

Does this have to be theoretical though?

Could you not just run loads of tests to see how far a certain stretch of bungee relates to projectile range and end up with an emipirical formula? More fun and easier then slogging away with theory that might not represent your system that accurately...

Of course for completeness you could do both, and compare the two different methods...

Sorry!
#4
Jun10-08, 04:21 PM
P: 571
Catapult Experiment

yeah it was pretty fun and its not supposed to be trial and error the teacher said you'll lose marks if you do that hmmph. :(
Sorry!
#5
Jun10-08, 05:41 PM
P: 571
hmm well i am calculating the elastic constant using forces


i measured it at rest to be like7 1/8 inches and when i put a 4 kg paint can on it it stretched to 9 1/4 inches.

i used
Fe=Fg
kx=mg
k=mg/x

converted my units and i end up getting around 733 is this number huge or is it just me? i've never worked with values this great in class so i'm not sure if i did it properly.. lol :O

anyways if i did do it properly my general equation to find out how far it will go using energy should be
Ee=Ek + Eg

[tex]2(kx^2/2-mgh)/m=v^2[/tex]

Then find out V at the moment of launch and i can find how far it will go correct??? hmm :D was thinking this would be rediculously hard lol. hopefully i did it correctly. :P


NOTE*** I am not sure how to calculate how FAR to pull back the catapult though because the cord gets bent and the 'x' value assumes its stretched linear ah garbage.
tiny-tim
#6
Jun10-08, 05:57 PM
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Hi Sorry!
Quote Quote by Sorry! View Post
[tex]2(kx^2/2-mgh)/m=v^2[/tex]
What is h is it a sort of x.costheta?
Then find out V at the moment of launch and i can find how far it will go correct???
That's it!
NOTE*** I am not sure how to calculate how FAR to pull back the catapult though because the cord gets bent and the 'x' value assumes its stretched linear
I don't think the angle matters you're using energy, which is a scalar, and depends only on the length.
Sorry!
#7
Jun10-08, 11:19 PM
P: 571
h is the height which it leaves the catapult since not all the energy is put into kinetic energy some is lost to gravity to get it to that new height right? and for that equation it assumes it starts off at h=0 guess i should ADD that into the equation as that will determine how far back i need to pull the catapult. hmmmmm
tiny-tim
#8
Jun11-08, 03:16 AM
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Quote Quote by Sorry! View Post
h is the height which it leaves the catapult since not all the energy is put into kinetic energy some is lost to gravity to get it to that new height right? and for that equation it assumes it starts off at h=0 guess i should ADD that into the equation as that will determine how far back i need to pull the catapult. hmmmmm
hmm probably best to put h = 0 at the point where the bungee is tight but not stretched.

Then the h in the equation will be (minus) the stretching times sin of the angle, which is probably small enough to be negligible.


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